Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long a[110][110];
long long n, m;
long long temp[110][110];
long long cnt;
bool test(long long val) {
  cnt = a[0][0];
  long long x = val;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      temp[i][j] = a[i][j];
    }
  }
  for (int j = 0; j < m; j++) {
    if (temp[0][j] < x) {
      return false;
    } else {
      temp[0][j] -= x;
    }
  }
  for (int i = 1; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (temp[i][j] >= temp[0][j]) {
        temp[i][j] -= temp[0][j];
      } else {
        return false;
      }
    }
    for (int j = 1; j < m; j++) {
      if (temp[i][j] != temp[i][j - 1]) {
        return false;
      }
    }
  }
  for (int i = 1; i < n; i++) {
    cnt += temp[i][0];
  }
  for (int i = 1; i < m; i++) {
    cnt += temp[0][i];
  }
  return true;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> a[i][j];
    }
  }
  long long val = a[0][0];
  bool ans = false;
  long long ma = 1e18;
  long long in = -1;
  for (int i = 0; i <= val; i++) {
    if (test(i) == true) {
      if (cnt < ma) {
        ma = cnt;
        in = i;
      }
    }
  }
  if (in >= 0) {
    test(in);
    long long cnt2 = a[0][0];
    for (int i = 1; i < n; i++) {
      if (temp[i][0]) cnt2 += temp[i][0];
    }
    for (int i = 1; i < m; i++) {
      cnt2 += temp[0][i];
    }
    cout << cnt2 << endl;
    long long ff = temp[0][0];
    temp[0][0] = a[0][0] - ff;
    for (int i = 0; i < n; i++) {
      while (temp[i][0] != 0) {
        cout << "row " << i + 1 << endl;
        temp[i][0]--;
      }
    }
    temp[0][0] = ff;
    for (int i = 0; i < m; i++) {
      while (temp[0][i] != 0) {
        cout << "col " << i + 1 << endl;
        temp[0][i]--;
      }
    }
  } else {
    cout << -1 << endl;
  }
}